Question

Trigonometry. I can't figure it out. ​

Answer 1

You also seem to be given that
`m\angle ADC=50^\circ`, according to the diagram.

In triangle ABC, we have

`\cos\angle CAB=\frac39=\frac13\implies m\angle CAB=m\angle CAD=\cos^(-1)\frac13`

By the law of sines,

`(\sin\angle ADC)/(AC)=(\sin\angle CAD)/(CD)`

`\implies CD=\frac{(9\,\mathrm{cm})\sin\left(\cos^(-1)\frac13\right)}{\sin50^\circ}\approx11.1\,\mathrm{cm}`